Solutions to Practice Problems
Practice Problem 20.1
Physics dictates that antenna length is intrinsically tied to the wavelength of the signal it is transmitting or receiving. To
transmit a signal through the atmosphere with an antenna efficiently, the length of the antenna must be at least a tenth of a
wavelength long.
What is the approximate length of the antenna required to transmit the sound of a tuning fork (which creates musical note A =
440 Hz)? Note: this sound must be transduced into an electrical signal first before it is transmitted.

m
c 3 108 sec

 681.8 km so
f
440 Hz

10
 68.18 km
The antenna must be at least 68.18 km long.
Practice Problem 20.2
(a) What is the wavelength of an AM radio station whose transmission frequency is 1430 kHz?

m
3 108 sec
c

 209.8 m
f 1430 103 Hz
(b) What is the approximate antenna length if the station uses an antenna that is half the wavelength long?

2
 104.9 m
Practice Problem 20.3
If a carrier signal vc(t) = 9 sin(2π5000t) Volts is modulated by a sine wave vm(t) =7.5 sin(2π440t) V, what is the percentage
modulation of the resulting AM signal?
m = Vm / Vc = 7.5 / 9 = 0.833
m(%) = m x 100% = 83.3%
Practice Problem 20.4
Suppose we want to transmit the sound of a two chime doorbell (f1=349 Hz, f2= 440 Hz) using VLF (very low frequency)
communications (let fc = 20 kHz). Each of the chimes has an amplitude of 10V, and the carrier’s amplitude is 20V. Sketch the
frequency domain representation of the transmitted signal and determine the bandwidth.
Which of the two chime frequencies determines the bandwidth?
The transmitted signal contains the carrier, and a pair of sidebands for each tone in the information signal, so the
frequencies present are: 20 kHz, 20 kHz + 349 Hz = 20.349 kHz, 20 kHz – 349 Hz = 19.651 kHz, 20 kHz + 440 Hz =
20.440 kHz, and 20 kHz – 440 Hz = 19.56 kHz.
The height of the spike for the carrier frequency is at Vc = 20 V. The height of the spikes for each of the sidebands is
½ of the chime amplitudes (Vm/2 = 5 V).
The bandwidth is given by BW = 2 fmax = 2 (440 Hz) = 880 Hz, or is also the difference between the highest transmitted
frequency and the lowest transmitted frequency, 20.440 kHz – 19.56 kHz = 880 Hz. Since fmax = 440 Hz, it is the 440
Hz tone that determines the bandwidth.
Practice Problem 20.5
If a carrier signal
vm (t )  20sin  2 5000t  Volts is amplitude modulated by information signal
vm (t )  4sin  2 200t   6cos  2 400t  411  , sketch the frequency plot for the resulting AM signal and calculate the
transmission bandwidth.
The transmitted signal contains the carrier, and a pair of sidebands for each tone in the information signal, so the
frequencies present are: 5 kHz, 5 kHz + 200 Hz = 5.2 kHz, 5 kHz – 200 Hz = 4.8 kHz, 5 kHz + 400 Hz = 5.4 kHz, and 5
kHz – 400 Hz = 4.6 kHz.
The height of the spike for the carrier frequency is at Vc = 20 V. The height of the spikes for each of the sidebands is
½ of the informational tone amplitudes (for the 200 Hz tone, V1/2 = 2 V and for the 400 Hz tone, V2/2 = 3 V).
The bandwidth is given by BW = 2 fmax = 2 (400 Hz) = 800 Hz, or is also the difference between the highest transmitted
frequency and the lowest transmitted frequency, 5.4 kHz – 4.6 kHz = 800 Hz.